H​2 ​​ and O2 ​​ react to produce H​2O. What is the mass of H2​​O produced when 0.73 g of H​2 ​​ completely reacts with 3.28 g of O2 ​​ ?

Answer: The weight of the mouse on the Moon is 0.036 N

Explanation:

Weight is defined as the force exerted by the body on any surface. It is also defined as the product of mass of the body multiplied by the acceleration due to gravity.

Mathematically,

[tex]W=mg[/tex]

where,

W = weight of the mouse

m = mass of the mouse = [tex]22g=0.022kg[/tex]

g = acceleration due to gravity on moon = [tex]1.625m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=0.022kg\times 1.625 m/s^2=0.036N[/tex]

Hence, the weight of the mouse on the Moon is 0.036 N

After a balanced neutralization reaction between Sr(OH)2 and HNO3, the unused Sr(OH)2 determines the solution is basic.

To address this question, let's first write down the balanced equation for the reaction between strontium hydroxide (Sr(OH)2) and nitric acid (HNO3) which is:

Sr(OH)2 (aq) + 2HNO3 (aq) → Sr(NO3)2 (aq) + 2H2O (l)

Now let's calculate the moles of HNO3 added:

Volume of HNO3 = 55.0 mL = 0.055 L

Concentration of HNO3 = 0.200 M

Moles of HNO3 = Volume × Concentration = 0.055 L × 0.200 M = 0.011 mol

Next, calculate the moles of Sr(OH)2:

Mass of Sr(OH)2 = 15.0 g

Molar mass of Sr(OH)2 = 121.63 g/mol (approximately)

Moles of Sr(OH)2 = Mass ÷ Molar mass = 15.0 g ÷ 121.63 g/mol = 0.123 mol

According to the stoichiometry of the balanced equation, we need twice as many moles of HNO3 to react completely with Sr(OH)2. In this scenario, we have excess Sr(OH)2 (0.123 mol) compared to HNO3 (0.011 mol). Hence, all of the HNO3 will react, leaving some Sr(OH)2 unreacted.

After reaction, moles of Sr(OH)2 remaining = 0.123 mol - (0.011 mol × 1/2) = 0.1175 mol

The concentration of remaining Sr(OH)2 can be calculated by assuming the final volume is the sum of the volumes of the solutions mixed, which is an approximation for dilute solutions.

Since all the HNO3 has reacted, the resulting solution will be basic due to the excess Sr(OH)2 remaining.

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(a) The balanced equation for the reaction between the solutes is: [tex]\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Sr(NO}_3\text{)}_2(aq) + 2\text{H}_2\text{O}(l)[/tex]. (b) The concentrations of ions are: Sr²⁺ is 2.136 M, NO₃⁻ is 0.2 M, and OH⁻ is 4.272 M. (c) Resultant solution is basic in nature.

To solve the given problem, let's follow the steps one by one:

(a) Write a balanced equation for the reaction that occurs between the solutes.

Strontium hydroxide Sr(OH)₂ reacts with nitric acid HNO₃ to form strontium nitrate Sr(NO₃)₂ and water:

[tex]\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Sr(NO}_3\text{)}_2(aq) + 2\text{H}_2\text{O}(l)[/tex]

(b) Calculate the concentration of each ion remaining in solution.

1. Determine the moles of each reactant:

- Moles of Sr(OH)₂:

[tex]\text{Molar mass of Sr(OH)}_2 = 87.62 \, (\text{Sr}) + 2 \times 16.00 \, (\text{O}) + 2 \times 1.01 \, (\text{H}) = 121.64 \, \text{g/mol} \\\\\text{Moles of Sr(OH)}_2 = \frac{15.0 \, \text{g}}{121.64 \, \text{g/mol}} \approx 0.123 \, \text{mol}[/tex]

- Moles of HNO₃:

[tex]\text{Molarity of HNO}_3 = 0.200 \, \text{M} \\\\\text{Volume of HNO}_3 = 55.0 \, \text{mL} = 0.055 \, \text{L} \\\\\text{Moles of HNO}_3 = 0.200 \, \text{mol/L} \times 0.055 \, \text{L} = 0.011 \, \text{mol}[/tex]

2. Determine the limiting reactant:

According to the balanced equation, 1 mole of Sr(OH)₂ reacts with 2 moles of HNO₃. Therefore, the reaction requires:

[tex]\text{Moles of HNO}_3 \text{ required} = 0.123 \, \text{mol Sr(OH)}_2 \times 2 = 0.246 \, \text{mol}[/tex]

Since we only have 0.011 moles of HNO₃, it is the limiting reactant.

3. Calculate the remaining moles of Sr(OH)₂:

[tex]\text{Moles of Sr(OH)}_2 \text{ reacted} = \frac{0.011 \, \text{mol HNO}_3}{2} = 0.0055 \, \text{mol} \\\\\text{Remaining moles of Sr(OH)}_2 = 0.123 \, \text{mol} - 0.0055 \, \text{mol} = 0.1175 \, \text{mol}[/tex]

4. Determine the concentrations of ions in the solution:

- Volume of the final solution = volume of HNO₃ + volume of water from Sr(OH)₂ dissolution (approximately equal to volume of water added):

Assuming the solution volume remains approximately 55.0 mL + a negligible volume from Sr(OH)₂, the final volume is roughly  0.055 L.

- Concentration of Sr²⁺ ions:

Since only 0.0055 moles of Sr(OH)₂ reacted to form Sr(NO₃)₂, 0.1175 moles of Sr(OH)₂ remain, giving the concentration of Sr²⁺:

[tex]\text{Concentration of Sr}^{2+} = \frac{0.1175 \, \text{mol}}{0.055 \, \text{L}} \approx 2.136 \, \text{M}[/tex]

- Concentration of NO₃⁻ ions:

All HNO₃ dissociates, producing:

[tex]\text{Concentration of NO}_3^{-} = \frac{0.011 \, \text{mol}}{0.055 \, \text{L}} = 0.2 \, \text{M}[/tex]

- Concentration of OH⁻ ions:

From Sr(OH)₂, OH⁻ concentration:

[tex]\text{OH}^{-} \text{ from Sr(OH)}_2: \text{2 moles OH}^{-}\text{ per mole of Sr(OH)}_2 \\\\\text{Concentration of OH}^{-} = 2 \times \frac{0.1175 \, \text{mol}}{0.055 \, \text{L}} \approx 4.272 \, \text{M}[/tex]

(c) Is the resultant solution acidic or basic?

To determine if the solution is acidic or basic, we compare the concentrations of H⁺ and OH⁻. Since HNO₃ (a strong acid) was completely neutralized and excess Sr(OH)₂ (a strong base) remains, the solution will be basic due to the presence of significant OH⁻ ions.

Summary

- Balanced equation:

[tex]\text{Sr(OH)}_2(aq) + 2\text{HNO}_3(aq) \rightarrow \text{Sr(NO}_3\text{)}_2(aq) + 2\text{H}_2\text{O}(l)[/tex]

- Concentrations of ions in solution:

[tex]\text{Sr}^{2+}: 2.136 \, \text{M} \\\\ \text{NO}_3^{-}: 0.2 \, \text{M} \\\\ \text{OH}^{-}: 4.272 \, \text{M} \\\\[/tex]

- Resultant solution is basic due to the presence of excess OH⁻ ions.

A helium blimp containing 535 kg of helium has approximately 8.05 × 10¹⁸ helium atoms, calculated by dividing the mass by the molar mass of helium and then multiplying by Avogadro's number.

To determine how many helium atoms are in a helium blimp containing 535 kg of helium, we need to use Avogadro's number and the molar mass of helium. First, we find the number of moles of helium:

Mass of helium (m) = 535 kg = 535,000 g
Molar mass of helium (M) = 4.0026 g/mol

Number of moles (n) = m / M = 535,000 g / 4.0026 g/mol ≈ 133786.85 mol

Avogadro's number (N₀) = 6.02214076 × 10²³ atoms/mol

Number of helium atoms = n × N₀ ≈ 133786.85 mol × 6.02214076 × 10²³ atoms/mol ≈ 8.05 × 10²⁸ atoms

Therefore, a helium blimp containing 535 kg of helium has approximately 8.05 × 10²⁸ helium atoms.

The diaphragm can be adjusted during deep breathing exercises, forced breathing, and activities that require stabilizing the abdominal cavity's volume and pressure.

The diaphragm is a key muscle involved in breathing. It separates the thoracic and abdominal cavities and plays a vital role in the expansion and contraction of the thoracic cavity. The diaphragm contracts during inhalation to increase the volume of the thoracic cavity, allowing air to enter the lungs.

Conversely, it relaxes during exhalation to decrease the volume of the thoracic cavity and expel air from the lungs. The conditions under which you would adjust the diaphragm are: During deep breathing exercises, such as diaphragmatic breathing, where you consciously focus on contracting and relaxing the diaphragm to improve lung function.

During forced breathing or hyperpnea, which occurs during activities like exercise or singing, where the diaphragm and other accessory muscles contract to enhance breathing.

During situations where you need to stabilize the volume and pressure of the abdominal cavity, such as in activities like defecation, urination, or childbirth, which involve cooperation between the diaphragm and abdominal muscles.

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The diaphragm is a muscle that contracts and relaxes to allow breathing. It plays a key role in the process of inspiration, where it moves downward to expand the chest and allow air to enter the lungs.

The diaphragm is a dome-shaped, muscular partition separating the thoracic and abdominal cavities in mammals. It plays a crucial role in breathing by contracting and relaxing to change thoracic volume. This action creates pressure variations, facilitating inhalation and exhalation. The diaphragm is essential for respiratory function and overall physiological equilibrium.

The diaphragm is a large, dome-shaped muscle below the lungs that allows breathing to occur when it alternately contracts and relaxes. It plays a crucial role in the process of inspiration, where the diaphragm contracts and moves downward, causing the chest to expand and allowing air to flow into the lungs.

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3H2 + N2 ........> 2NH3

This means that each 6 grams of hydrogen react with 28 grams of nitrogen. To know how many grams of nitrogen are required to react with 2 grams of hydrogen, we will simply do cross multiplication as follows:

mass of nitrogen = (2 x 28) / 6 = 9.334 grams

Therefore, if we have 11.3 grams of nitrogen, 9.334 grams would react with 2 grams of hydrogen.

remaining mass of nitrogen = 11.3 - 9.334 = 1.966 grams

Final answer:

When magnesium ribbon burns with oxygen, it forms a bright, white light and a white, powdery product called magnesium oxide (MgO).

Explanation:

When magnesium ribbon burns with oxygen, it forms a bright, white light and a white, powdery product called magnesium oxide (MgO). The chemical equation for this reaction is:

2Mg (s) + O₂(g) → 2MgO (s)

This is an example of a combination reaction where an element (magnesium) combines with oxygen to form an oxide. The reaction is accompanied by the release of heat and light.

Answer : The molarity of vitamin c in organic juice is, [tex]2.97\times 10^{-3}mole/L[/tex]

Solution : Given,

Mass of ascorbic acid (solute) = 124 mg = 0.124 g        [tex](1mg=0.001g)[/tex]

Volume of juice = 236.6 ml

Molar mass of ascorbic acid = 176.12 g/mole

Formula used :

[tex]Molarity=\frac{\text{Moles of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}[/tex]

Now put all the given values in this formula, we get the molarity of vitamin c in organic juice.

[tex]Molarity=\frac{0.124g\times 1000}{176.12g/mole\times 236.6ml}=2.97\times 10^{-3}mole/L[/tex]

Therefore, the molarity of vitamin c in organic juice is, [tex]2.97\times 10^{-3}mole/L[/tex]

Final answer:

The molarity of vitamin C in orange juice, where one cup contains 124 mg of ascorbic acid and is equal to 236.6 mL, is approximately 0.00297 M.

Explanation:

To calculate the molarity of vitamin C in orange juice, we can use the formula for molarity:

Molarity = moles of solute / liters of solution

First, we need to find the number of moles of ascorbic acid in one cup of orange juice.

1 cup of orange juice contains 124 mg of ascorbic acid. Since the molar mass of ascorbic acid, C6H8O6, is 176.12 g/mol, we convert the mass from mg to grams and then to moles:

124 mg * (1 g / 1000 mg) * (1 mol / 176.12 g) = 0.000704 mol

Next, we convert the volume from milliliters to liters:

236.6 mL * (1 L / 1000 mL) = 0.2366 L

Now, we can calculate the molarity:

Molarity = 0.000704 mol / 0.2366 L ≈ 0.00297 M

Thus, the molarity of vitamin C in the orange juice is approximately 0.00297 M.

Answer:  C.  Place the coefficient 2 in front of elemental iron.

Explanation:   The given chemical reaction is

                 3CO(g) + Fe2O3(s) → Fe(s) + 3CO2(g)

As we can see that stiochiometric coefficient of Carbon is balanced and oxygen coefficient is also balanced. The atom left with unbalanced coefficient is Fe.

Thus By Placing the coefficient 2 in front of the elemental iron will lead to the balanced equation.

Thus the balanced equation can be written as -

                 3CO(g) + Fe2O3(s) → 2Fe(s) + 3CO2(g)

Answer is: the reducing agent is C₂H₅OH.

Balanced chemical reaction:

16H⁺ + 2Cr₂O₇²⁻ + C₂H₅OH → 4Cr³⁺ + 11H₂O + 2CO₂.

Oxidation half reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O/ ×2.

2Cr₂O₇²⁻ + 28H⁺ + 12e⁻ → 4Cr³⁺ + 14H₂O.

Reduction half reaction: C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻.

Net reaction: 2Cr₂O₇²⁻ + 28H⁺ + C₂H₅OH + 3H₂O → 4Cr³⁺ + 14H₂O + 2CO₂ + 12H⁺.

Reducing agent is element or compound who loose electrons in chemical reaction. Ethanol (C₂H₅OH) lost electrons and it is oxidized to carbon dioxide (CO₂).

The reducing agent in the reaction is [tex]\boxed{\text{C}_{2}\text{H}_{5}\text{OH}}[/tex].

Further Explanation:

Redox reaction:

Redox is a term that is used collectively for the reduction-oxidation reaction. It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out simultaneously. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The process of gain of electrons or the decrease in the oxidation state of the atom is called reduction while that of loss of electrons or the increase in the oxidation number is known as oxidation. In redox reactions, one species lose electrons and the other species gain electrons. The species that lose electrons and itself gets oxidized is called as a reductant or reducing agent. The species that gains electrons and gets reduced is known as oxidant or oxidizing agent. The presence of redox pair or redox couple is a must for the redox reaction.

The general representation of a redox reaction is,

[tex]\text{X}+\text{Y}\rightarrow\text{X}^{+}+\text{Y}^{-}[/tex]

The oxidation half-reaction can be written as:

[tex]\text{X}\rightarrow\text{X}^{+}+e^{-}[/tex]  

The reduction half-reaction can be written as:

[tex]\text{Y}+e^{-}\rightarrow\text{Y}^{-}[/tex]

Here, X is getting oxidized and its oxidation state changes from  to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1. Hence, X acts as the reducing agent, whereas Y is an oxidizing agent.

Rules to calculate the oxidation states of elements:

1. The oxidation number of free element is always zero.

2. The oxidation number of oxygen is generally taken as -2, except for peroxides.

3. The oxidation state of hydrogen is normally taken as +1.

4. The sum of oxidation numbers of all the elements present in a neutral compound is zero.

5. The oxidation numbers of group 1 and group 2 elements are +1 and +2 respectively.

The oxidation state of O is -2.

The expression to calculate the oxidation number in [tex]\text{Cr}_{2}\text{O}_{7}^{2-}[/tex] is:

[tex]\left[2(\text{Oxidation state of Cr})+7(\text{oxidation state of O})\right]=-2[/tex]     ...... (1)

Rearrange equation (1) for oxidation number of Cr

[tex]2(\text{Oxidation number of Cr})=\left[(-2)-7(\text{oxidation number of O})\right][/tex]                  …… (2)

Substitute -2 for oxidation state of O in equation (2).

[tex]\begin{aligned}\text{Oxidation state of Cr}&=\dfrac{[(-2)-7(-2)]}{2}\\&=\dfrac{[-2+14]}{2}\\&=+6\end{aligned}[/tex]

The charge on Cr is +3 so its oxidation state is also +3. The oxidation state of Cr changes from +6 to +3. This shows it decreases during the reaction and therefore Cr is an oxidizing agent.

The oxidation state of O is -2 and the oxidation state of H is +1.

The expression to calculate the oxidation in [tex]\text{C}_{2}\text{H}_{5}\text{OH}[/tex] is:

[tex]\left[2(\text{oxidation state of C})+1(oxidation state of O)+6(oxidation state of H)\right]=0[/tex]   ...... (3)

Rearrange equation (3) for oxidation state of C

[tex]2(\text{oxidation state of C})=\left[-1(oxidation state of O)-6(oxidation state of H)\right][/tex]       ...... (4)

Substitute -2 for oxidation state of O and +1 for oxidation state of H in equation (4).

[tex]\begin{aligned}\text{Oxidation state of C}&=\dfrac{[-1(-2)-6(+1)]}{2}\\&=\dfrac{[+2-6]}{2}\\&=-2\end{aligned}[/tex]

The oxidation state of O is -2.

The expression to calculate the oxidation state in [tex]\text{CO}_{2}[/tex] is:

[tex]\left[(\text{oxidation state of C})+2(oxidation state of O)\right]=0[/tex]   ...... (5)

Rearrange equation (5) for oxidation state of C

[tex](\text{oxidation state of C})=\left[-2(oxidation state of O)\right][/tex]       ...... (6)

Substitute -2 for oxidation state of O in equation (6).

[tex]\begin{aligned}\text{Oxidation state of C}&=[-2(-2)]\\&=+4\end{aligned}[/tex]

The oxidation state of C changes from -2 to +4. This shows it increases during the reaction and therefore [tex]\textbf{C}_{2}\textbf{H}_{5}\textbf{OH}[/tex] acts as a reducing agent.

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Answer details:

Grade: High School

Subject: Chemistry

Chapter: Redox reactions

Keywords: redox reaction, C2H5OH, Cr, C, CO2, Cr2O72-, oxidation, reduction, reductant, oxidant, reducing agent, oxidizing agent, electrons, redox pair, redox couple, oxidation state, oxidized, reduced, simultaneously.

Answer: Option (D) is the correct answer.

Explanation:

When big rocks are broken down into smaller rocks due to the natural processes then this process of breaking is known as mechanical weathering.

Mechanical weathering is also known as physical weathering. It arises due to physical forces caused by rainwater and change in temperature etc.

Thus, we can conclude that out of the given options, physical forces is the cause of mechanical weathering.

Answer:

A. Tools are appropriate for the children in your setting and meet professional requirements for quality.

Explanation:

This is correct! C::

Atomic Number is the same as the number of protons in an element.
Mass Number is the number of Protons + Neutrons in an element.

Atomic Number: 6 means 6 Protons
Mass Number: 15 means 15 atoms that are a proton/neutron.
We are given out of the 15 atoms, 6 of them are protons, so the other 9 must be Neutrons.

15 - 6 = 9 so there must be 9 Neutrons.

There are 9 Neutrons in atom's nucleus.

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The percentage by mass of calcium carbonate present in rock is [tex]\boxed{{\mathbf{77}}{\mathbf{.62 \% }}}[/tex].

Further Explanation:

First, we have to find the excess number of moles of HCl acid that are neutralized by NaOH.

The number of moles of NaOH in 11.56 ml of 1.010 M NaOH solution is calculated as follows:

[tex]\begin{aligned}{\text{Number of moles of NaOH}}\left({{\text{mol}}}\right)&={\text{Concentration }}\left( {{\text{mol/L}}}\right) \times{\text{Volume }}\left({\text{L}} \right)\\&= 1.010{\text{ mol/L}}\left({11.56{\text{ ml}}\times \frac{{1{\text{ L}}}}{{1000{\text{ml}}}}}\right)\\&=0.011676{\text{ mol}}\\\end{aligned}[/tex]

The balanced chemical reaction of NaOH and HCl is as follows:

[tex]{\text{NaOH}}\left({aq}\right)+{\text{HCl}}\left({aq}\right)\to{\text{NaCl}}\left({aq} \right) +{{\text{H}}_{\text{2}}}{\text{O}}\left( l \right)[/tex]

Since NaOH and HCl are reacted in 1:1 ratio, therefore, the excess number of moles of HCl are equal to number of moles of NaOH that is 0.011676 mol.

Now, we have to find how many moles of HCl initially reacted with limestone.

The initial number of moles of HCl in 30.00 ml of 1.035 M HCl solution is calculated as follows:

[tex]\begin{aligned}{\text{Number of moles of HCl}}\left( {{\text{mol}}}\right)&={\text{Concentration }}\left( {{\text{mol/L}}}\right) \times {\text{Volume }}\left({\text{L}} \right)\\&= 1.035{\text{ mol/L}} \times \left( {30.00{\text{ ml}}\times \frac{{1{\text{ L}}}}{{1000\,{\text{ml}}}}}\right)\\&=0.03105{\text{ mol}}\\\end{aligned}[/tex]

Therefore, the number of moles of HCl initially reacted with limestone is calculated as follows:

[tex]\begin{aligned}{\text{Number of moles of HCl reacted with CaC}}{{\text{O}}_3}&=\left({0.03105{\text{ mol}} - 0.011676{\text{ mol}}}\right)\\&={\text{0}}{\text{.019374 mol of HCl}}\\\end{aligned}[/tex]

Therefore, the number of moles of HCl initially reacted with limestone is 0.019374 mol.

The balanced chemical equation for the reaction of limestone [tex]\left( {{\text{CaC}}{{\text{O}}_{\text{3}}}}\right)[/tex] with HCl is as follows:

[tex]{\text{CaC}}{{\text{O}}_3}\left( s \right) + 2{\text{HCl}}\left( {aq} \right)\to{\text{C}}{{\text{O}}_{\text{2}}}\left( g \right) + {{\text{H}}_{\text{2}}}{\text{O}}\left( l \right) + {\text{CaC}}{{\text{l}}_2}\left( {aq} \right)[/tex]

The balanced chemical equation shows that 1 mole of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] reacted with 2 moles of HCl to neutralize the reaction completely, therefore, the number of moles of  [tex]{\text{CaC}}{{\text{O}}_3}[/tex] neutralized by 0.019374 mol of HCl are calculated as follows:

[tex]\begin{aligned}{\text{Amount of CaC}}{{\text{O}}_3}\left( {{\text{mol}}}\right)&= \left( {{\text{0}}{\text{.019374 mol of HCl}}}\right)\left({\frac{{1{\text{ mol CaC}}{{\text{O}}_3}}}{{{\text{2 mol HCl}}}}}\right)\\&=0.009687{\text{ mol CaC}}{{\text{O}}_3}\\\end{aligned}[/tex]

The molar mass of [tex]{\text{CaC}}{{\text{O}}_3}[/tex]  is 100.0 g/mol.

Mass of 0.009687 mol of [tex]{\text{CaC}}{{\text{O}}_3}[/tex] is calculated as follows:

[tex]\begin{aligned}{\text{Mass}}\left( {\text{g}}\right)&={\text{Number of moles}} \times{\text{Molarmass}}\left({{\text{g/mol}}}\right)\\&=0.009687{\text{mol}}\times{\text{100}}{\text{.0 g/mol}}\\&=0.9687{\text{g}}\\\end{aligned}[/tex]

The percentage by mass can be calculated as follows:

[tex]\begin{aligned}{\text{Percent by mass}}\left( \%\right)&=\frac{{{\text{Mass of CaC}}{{\text{O}}_3}}}{{{\text{Mass of lime stone}}}}\times 100\\&=\frac{{0.9687{\text{ g}}}}{{1.248{\text{ g}}}}\times 100\\&=77.62{\text{ }}\%\\\end{aligned}[/tex]

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Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: Percentage by mass, calcium carbonate in rock, number of moles of HCl, excess number of moles, CaCO3, balance chemical equation, limestone.

Final answer:

To calculate the percent by mass of calcium carbonate in the limestone, we calculate the moles of HCl that reacted with the limestone, subtract the moles neutralized by NaOH, convert this to grams of CaCO₃, and divide by the sample mass.

Explanation:

To calculate the percent by mass of calcium carbonate in the limestone rock, we need to find out how much of the HCl was used to react with the calcium carbonate and not neutralized by NaOH, and then convert this amount to grams of CaCO3.

We start by calculating the number of moles of NaOH that reacted with the excess HCl:

Number of moles of NaOH = Volume (L) × Molarity (M)

Number of moles of NaOH = 0.01156 L × 1.010 M = 0.0116776 mol

Since the reaction between NaOH and HCl is 1:1, the moles of HCl that were neutralized by NaOH are also 0.0116776 mol.

Now we can calculate the moles of HCl that reacted with the CaCO3:

Total moles of HCl initially = Volume (L) × Molarity (M)

Total moles of HCl = 0.03000 L × 1.035 M = 0.03105 mol

Moles of HCl that reacted with CaCO₃ = Total moles of HCl - Moles of HCl neutralized by NaOH

Moles of HCl that reacted with CaCO₃ = 0.03105 mol - 0.0116776 mol = 0.0193724 mol

The reaction between CaCO₃ and HCl is also 1:1:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Moles of CaCO₃ = Moles of HCl that reacted with CaCO₃ = 0.0193724 mol

To find the mass of CaCO₃ we multiply the moles of CaCO₃ by the molar mass of CaCO₃ (100.09 g/mol):

Mass of CaCO₃ = Moles of CaCO₃ × Molar Mass of CaCO₃

Mass of CaCO₃ = 0.0193724 mol × 100.09 g/mol = 1.93936 g

Finally, we calculate the percent by mass of CaCO₃ in the rock:

Percent by mass of CaCO₃ = (Mass of CaCO₃ / Mass of Rock Sample) × 100%

Percent by mass of CaCO₃ = (1.93936 g / 1.248 g) × 100% = 155.413%

However, a percent mass over 100% indicates an error in the calculation as it's not possible to have more calcium carbonate than the total mass of the rock. It likely means that we must take into account other substances in the limestone that might react with HCl or a calculation error.

The solubility of nitrogen in a diver's blood at a pressure of 2.85 atm and a temperature of 25.0 ∘c is 1.78 x 10^-3 moles per litre, according to Henry's law.

In this context, we will use Henry's law to calculate the solubility of N2 in a diver's blood. According to Henry's law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is: C = P * k, where C is the solubility of the gas in the liquid, P is the partial pressure of the gas, and k is Henry's law constant.

In the given situation, the total pressure is 2.85 atm (which is the sum of the partial pressures of all gases), and the value of Henry's law constant (k) for N2 is 6.26×10−4mol/(l⋅atm). So, the solubility of nitrogen can be calculated by multiplying the pressure by Henry's law constant: C = 2.85 atm * 6.26×10−4mol/(l⋅atm) = 1.78 x 10^-3 mol/L.

Therefore, at a pressure of 2.85 atm and a temperature of 25.0 ∘c, the solubility of nitrogen in a diver's blood is 1.78 x 10^-3 moles per litre.

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The balanced chemical equation for the combustion of benzene is C6H6(g) + 15O2(g) → 6CO2(g) + 3H2O(g). AS is expected to be positive in this process.

The balanced chemical equation for the combustion of benzene, C6H6, is:

C6H6 (g) + 15O2 (g) → 6CO2 (g) + 3H2O (g)

In this process, AS (entropy change) is expected to be positive since the reaction produces more gas molecules (CO2 and H2O) than the reactant (benzene).

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Answer:

There are 8.43x10²¹ total atoms in 0.280 g of P₂O₅.

Explanation:

Let's follow some steps to calculate the total atoms.

1st) Calculate the molar mass of P₂O₅

Look for the atomic weight of each atom in the Periodic table:

- Atomic weight of Phosphorus = 31 g/mol

- Atomic weight of Oxygen = 16 g/mol

Then, multiply each atomic weight by its coefficient to calculate the molar mass of P₂O₅:

(Atomic weight of P .2) + (Atomic weight of O .5)= Molar mass of P₂O₅

(31 g/mol.2) + (16 g/mol.5) = 62 g/mol + 80 g/mol = 142 g/mol

2nd) Calculate the moles of P₂O₅ that are contained in 0.280 g

This step is easy using a Rule of three thinking that if 142 g of P₂O₅ represents a mol of P₂O₅ the 0.280 g will be "x" moles:

   142 g -------- 1 mol of P₂O₅

0.280 g -------- x = (0.280 g.1 mol)/142 g =0.002 mol of P₂O₅

This means that 0.002 moles of P₂O₅ weights 0.280g.

3rd) Calculate the moles of P and O

To see clear how many moles of each atoms are in the molecule of P₂O₅ we disassociate it:

                                         P₂O₅   →  2P  +  5O

From the reaction we know that 1 mol of P₂O₅ produces 2 moles of phosphorus and 5 moles of oxygen. Now we can make a relation and thinking that if 1 mol of P₂O₅ produces 2 moles of P the 0.002 moles of P₂O₅ that we have will produce "x" moles of P:

        1 mol of P₂O₅ ----- 2 moles of P

0.002 mol of P₂O₅ ----- x = (0.002 mol.2moles)/1mol = 0.004 moles of P

We use the same reasoning for oxygen:

        1 mol of P₂O₅ ------- 5 moles of O

0.002 mol of P₂O₅ ------- x = (0.002 mol.5moles)/1mol = 0.01 moles of O

Up to here we have 0.004 moles of atoms of phosphorus and 0.01 moles of atoms of oxygen.

4th) Calculate the total atoms

To this step it is important to remember that 1 mol of something representa a quantity of 6.022x10²³ (that is called Avogadro's number) So, 1 mol of atoms represents 6.022x10²³ atoms.

Now, if we know that in 1 mol of phosphorus atoms is equal to 6.022x10²³ atoms of phosphorus the 0.004 moles that we have will be equal to "x" atoms:

        1 mol of P ------ 6.022x10²³ atoms of P

0.004 mol of P ------ x = (0.004 . 6.022x10²³)/ 1 = 2.41x10²¹ atoms of P

Use the same reasoning for oxygen:

     1 mol of O ------ 6.022x10²³ atoms of O

0.01 mol of O ------ x = (0.01 . 6.022x10²³)/ 1 = 6.022x10²¹ atoms of O

Now that we have the number of atoms of phosphorus and oxygen let's sum them to find the total atoms:

2.41x10²¹ atoms of P + 6.022x10²¹ atoms of O = 8.43x10²¹ total atoms

Finally, there are 8.43x10²¹ total atoms (of phosphorus and oxygen) in 0.280 g of P₂O₅.

In the given question, Insulin, adrenaline, and estrogen are examples of hormones. Each hormone has a specific function and plays an important role in maintaining homeostasis in the body.

Insulin is a hormone made up of 51 amino acids and is released by the beta cells of pancreas. It helps in regulating blood glucose level.

Adrenaline is also known as epinephrine. It is produced by the adrenal glands and is involved in the body's "fight or flight" response to stress.

Estrogen is a female sex hormone produced by the ovaries and is involved in the development of female secondary sexual characteristics, such as breast development and the menstrual cycle.

Therefore, examples of hormones are insulin, adrenaline, and estrogen, which are chemical messengers that regulate many physiological processes in the body.

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Acetylene has a chemical formula which can be written as:

C2H2

We can see that there are two positive ions, H+. Now what deprotonation means is that the H+ is removed from acetylene to form acetylene ion and water. In this case, I believe that the answer would be:

LiOCH3

LiOCH₃ would not effectively deprotonate acetylene

The equilibrium reaction can be determined if the pKa or Ka values ​​of the acid and conjugate acids (acids in the product) are known.

So it can be concluded that the reacting acid can protonate the base or vice versa base compounds can deprotonating the acid in the reaction, so that the reaction can proceed to the right to form a product or not

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid

Acids and bases according to Bronsted-Lowry

If the acid gives (H⁺), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H⁺), then the base is formed can release protons and is called the conjugate acid from the original base.

The value of the equilibrium constant (K)

Can be formulated:

K acid-base reaction = Ka acid on the left / K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K

[tex]\large{\boxed{\bold{K\:=\:10^{-pK}}}}[/tex]

K value> 1) indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, the pKa value of the conjugated acid from the base of the compounds above

Whereas the pKa of Acetylene (C₂H₂) itself is = 25

From the conjugated acid pKa value of some of the bases above shows only LiOCH₃ bases that cannot deprotonate acetylene because the pKa value is smaller than the pKa acetylene

The reactions that occur are:

LiOCH₃ + HC ≡ CH ---> HOCH₃ + LiC = CH

The value of the equilibrium constant K is

pK = pKa acetylene - pKa HOCH₃

pK = 25-16

pK = 9

[tex]K\:=\:10^{-pK}[/tex]

[tex]K\:=\:10^{-9}[/tex]

K values ​​<1 indicate a reaction cannot occur

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Keywords : pKa,  acetylene, deprotonate, the conjugate acid